# Find the following limit. lim_{nrightarrowinfty}frac{2n^2+2}{5n^3-3n-1}

Question
Limits and continuity
Find the following limit.
$$\lim_{n\rightarrow\infty}\frac{2n^2+2}{5n^3-3n-1}$$

2020-11-11
Solution:
$$\lim_{n\rightarrow\infty}\frac{2n^2+2}{5n^3-3n-1}=\lim_{n\rightarrow\infty}\frac{n^2(2+\frac{2}{n^2})}{n^3(5-\frac{3}{n^2}-\frac{1}{n^3})}$$
$$=\lim_{n\rightarrow\infty}\frac{(2+\frac{2}{n^2})}{n(5-\frac{3}{n^2}-\frac{1}{n^3})}$$
$$=\lim_{n\rightarrow\infty}\frac{1}{n}\times\lim_{n\rightarrow\infty}\frac{(2+\frac{2}{n^2})}{(5-\frac{3}{n^2}-\frac{1}{n^3})}$$
$$=\lim_{n\rightarrow\infty}\frac{1}{n}\times\frac{(2+0)}{(5-0-0)}$$
$$=\lim_{n\rightarrow\infty}\frac{1}{n}\times\frac{2}{5}$$
$$=0\times\frac{2}{5}$$
$$=0$$

### Relevant Questions

Verify, using the definition of convergence of a sequence, that the following sequences converge to the proposed limit.
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